NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.3
Exercise 2.3 of Class 9 Maths Chapter 2 - Polynomials deals with finding the zeros of a polynomial. In this exercise, you will learn how to verify whether a number is a zero of a polynomial. This understanding of the zeros of a polynomial is necessary to learn about the values the polynomial takes, and ways to think about the numbers.
These questions and solutions are significant for Class 9 Maths and also in higher classes. The exercise is made according to the latest NCERT syllabus, covered with the current significant points needed to succeed in the CBSE examinations. The NCERT Solutions in Exercise 2.3 are written step by step to ensure that they are easy and understandable.
1.0Download NCERT Solutions of Class 9 Maths Chapter 2 Polynomials Exercise 2.3: Free PDF
The NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 helps you learn how to find if a number is a zero of a polynomial and explain each step clearly. Download the free PDF of the Solutions from below:
2.0NCERT Exercise Solutions Class 9 Chapter 2- Polynomials: All Exercises
3.0NCERT Class 9 Maths Chapter 2 Exercise 2.3 : Detailed Solutions
1. Determine which of the following polynomials has (x+1) a factor :
(i) x³+x²+x+1
(ii) x⁴+x³+x²+x+1
(iii) x⁴+3x³+3x²+x+1
(iv) x³−x²−(2+√2)x+√2
Sol. (Using Factor Theorem: (x+1) is a factor of p(x) if p(−1)=0)
(i) Let p(x)=x³+x²+x+1
The zero of x+1 is -1.
p(−1) = (−1)³+(−1)²+(−1)+1 = −1+1−1+1 = 0.
By Factor Theorem, x+1 is a factor of p(x).
(ii) Let p(x)=x⁴+x³+x²+x+1
The zero of x+1 is -1.
p(−1) = (−1)⁴+(−1)³+(−1)²+(−1)+1 = 1−1+1−1+1 = 1.
Since 1≠0, by Factor Theorem, x+1 is not a factor of p(x).
(iii) Let p(x)=x⁴+3x³+3x²+x+1
The zero of x+1 is -1.
p(−1) = (−1)⁴+3(−1)³+3(−1)²+(−1)+1 = 1−3(1)+3(1)−1+1 = 1−3+3−1+1 = 1.
Since 1≠0, by Factor Theorem, x+1 is not a factor of p(x).
(iv) Let p(x)=x³−x²−(2+√2)x+√2
The zero of x+1 is -1.
p(−1) = (−1)³−(−1)²−(2+√2)(−1)+√2
= −1−1 + (2+√2) + √2
= −2 + 2 + √2 + √2 = 2√2.
Since 2√2≠0, by Factor Theorem, x+1 is not a factor of p(x).
2. Use the factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases :
(i) p(x)=2x³+x²−2x−1, g(x)=x+1.
(ii) p(x)=x³+3x²+3x+1, g(x)=x+2.
(iii) p(x)=x³−4x²+x+6; g(x)=x−3.
Sol. (i) p(x)=2x³+x²−2x−1, g(x)=x+1.
Set g(x)=0 => x+1=0 => x=−1. Therefore, the zero of g(x) is -1.
Now, p(−1) = 2(−1)³+(−1)²−2(−1)−1 = 2(−1)+1+2−1 = −2+1+2−1 = 0.
By Factor Theorem, since p(−1)=0, g(x) is a factor of p(x).
(ii) Let p(x)=x³+3x²+3x+1, g(x)=x+2.
Set g(x)=0 => x+2=0 => x=−2. Therefore, the zero of g(x) is -2.
Now, p(−2) = (−2)³+3(−2)²+3(−2)+1 = −8+3(4)−6+1 = −8+12−6+1 = −14+13 = −1.
By Factor Theorem, since p(−2)≠0, g(x) is not a factor of p(x).
(iii) p(x)=x³−4x²+x+6, g(x)=x−3.
Set g(x)=0 => x−3=0 => x=3. Therefore, the zero of g(x)=3.
Now p(3) = (3)³−4(3)²+3+6 = 27−4(9)+3+6 = 27−36+3+6 = 36−36 = 0.
By Factor Theorem, since p(3)=0, g(x) is a factor of p(x).
4. Find the value of k, if x−1 is a factor of p(x) in each of the following cases :
(i) p(x)=x²+x+k
(ii) p(x)=2x²+kx+√2
(iii) p(x)=kx²−√2x+1
(iv) p(x)=kx²−3x+k
Sol. If (x−1) is a factor of p(x), then by Factor Theorem, p(1)=0.
(i) p(x)=x²+x+k
p(1) = (1)²+(1)+k = 0
1+1+k = 0
2+k = 0 => k=−2.
(ii) p(x)=2x²+kx+√2
p(1) = 2(1)²+k(1)+√2 = 0
2+k+√2 = 0 => k=−(2+√2).
(iii) p(x)=kx²−√2x+1
p(1) = k(1)²−√2(1)+1 = 0
k−√2+1 = 0 => k=√2−1.
(iv) p(x)=kx²−3x+k
p(1) = k(1)²−3(1)+k = 0
k−3+k = 0
2k−3 = 0 => 2k=3 => k=3/2.
5. Factorise :
(i) 12x²−7x+1
(ii) 2x²+7x+3
(iii) 6x²+5x−6
(iv) 3x²−x−4
Sol. (i) 12x²−7x+1 (Split the middle term: product = 12x² × 1 = 12x², sum = -7x. Factors: -4x, -3x)
= 12x²−4x−3x+1
= 4x(3x−1)−1(3x−1)
= (3x−1)(4x−1)
(ii) 2x²+7x+3 (Split the middle term: product = 2x² × 3 = 6x², sum = 7x. Factors: 6x, x)
= 2x²+6x+x+3
= 2x(x+3)+1(x+3)
= (x+3)(2x+1)
(iii) 6x²+5x−6 (Split the middle term: product = 6x² × (-6) = -36x², sum = 5x. Factors: 9x, -4x)
= 6x²+9x−4x−6
= 3x(2x+3)−2(2x+3)
= (3x−2)(2x+3)
(iv) 3x²−x−4 (Split the middle term: product = 3x² × (-4) = -12x², sum = -x. Factors: -4x, 3x)
= 3x²−4x+3x−4
= x(3x−4)+1(3x−4)
= (x+1)(3x−4)
6. Factorise : (using Factor Theorem and polynomial division/grouping)
(i) x³−2x²−x+2
(ii) x³−3x²−9x−5
(iii) x³+13x²+32x+20
(iv) 2y³+y²−2y−1
Sol. (i) x³−2x²−x+2
Let p(x)=x³−2x²−x+2
By trial, test integer factors of the constant term (2): ±1, ±2.
p(1) = (1)³−2(1)²−(1)+2 = 1−2−1+2 = 0.
Therefore, by Factor Theorem, (x−1) is a factor of p(x).
Now, perform polynomial division or grouping:
x³−2x²−x+2 = x²(x−1)−x(x−1)−2(x−1) (Factoring by grouping can also be x²(x-2)-(x-2) = (x²-1)(x-2))
= (x−1)(x²−x−2)
Now factor the quadratic: x²−x−2 = x²−2x+x−2 = x(x−2)+1(x−2) = (x−2)(x+1)
So, x³−2x²−x+2 = (x−1)(x−2)(x+1).
(ii) x³−3x²−9x−5
Let p(x)=x³−3x²−9x−5
By trial, test integer factors of the constant term (5): ±1, ±5.
p(−1) = (−1)³−3(−1)²−9(−1)−5 = −1−3(1)+9−5 = −1−3+9−5 = −4+9−5 = 0.
Therefore, by Factor Theorem, x=−1 or (x+1) is a factor of p(x).
Now, perform polynomial division or grouping:
x³−3x²−9x−5 = x²(x+1)−4x(x+1)−5(x+1) (Here, we strategically add/subtract terms to get (x+1) as common)
= (x+1)(x²−4x−5)
Now factor the quadratic: x²−4x−5 = x²−5x+x−5 = x(x−5)+1(x−5) = (x−5)(x+1)
So, x³−3x²−9x−5 = (x+1)(x+1)(x−5) = (x+1)²(x−5).
(iii) x³+13x²+32x+20
Let p(x)=x³+13x²+32x+20
By trial, test integer factors of the constant term (20): ±1, ±2, ±4, ±5, ±10, ±20.
p(−1) = (−1)³+13(−1)²+32(−1)+20 = −1+13(1)−32+20 = −1+13−32+20 = 33−33 = 0.
Therefore, by Factor Theorem, x=−1 or (x+1) is a factor of p(x).
Now, perform polynomial division or grouping:
x³+13x²+32x+20 = x²(x+1)+12x(x+1)+20(x+1)
= (x+1)(x²+12x+20)
Now factor the quadratic: x²+12x+20 = x²+2x+10x+20 = x(x+2)+10(x+2) = (x+2)(x+10)
So, x³+13x²+32x+20 = (x+1)(x+2)(x+10).
(iv) 2y³+y²−2y−1
Let p(y)=2y³+y²−2y−1
By trial, test integer factors of the constant term (-1) divided by factors of the leading coefficient (2): ±1, ±1/2.
p(1) = 2(1)³+(1)²−2(1)−1 = 2+1−2−1 = 0.
Therefore, by Factor Theorem, (y−1) is a factor of p(y).
Now, perform polynomial division or grouping:
2y³+y²−2y−1 = 2y²(y−1)+3y(y−1)+1(y−1)
= (y−1)(2y²+3y+1)
Now factor the quadratic: 2y²+3y+1 = 2y²+2y+y+1 = 2y(y+1)+1(y+1) = (y+1)(2y+1)
So, 2y³+y²−2y−1 = (y−1)(y+1)(2y+1).
4.0Key Features and Benefits Class 9 Maths Chapter 2 Number System: Exercise 2.3
- Exercise 2.3 teaches how to check whether a number is a zero of a polynomial.
- The questions conform to the NCERT syllabus as per the latest regulations and also follow the CBSE examination pattern.
- This topic will be the foundation to some of the algebra topics that you will learn in future chapters.
- NCERT solutions keep you to the point, with clear step by step solutions that help in clear understanding and avoiding confusions.
- Regularly practicing these exercises can help you improve your confidence and your problem solving abilities for your school tests and different competitive exams like the olympiads.
